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A balloon filled with 5.44 L of helium initially at 2.46 atm of pressure rises into the atmosphere. When the surrounding pressure reaches 440.0 mmHg, the balloon will burst. If 1 atm = 760 mmHg, the volume of the balloon at bursting point is:

a. 7.26 L
b. 11.08 L
c. 9.92 L
d. 5.44 L

1 Answer

3 votes

Final answer:

Using the combined gas law and converting pressures to the same units, the balloon will burst at a volume of approximately 23.11 L, which is not one of the provided options. There may be an error in the question or the options.

Step-by-step explanation:

The volume of the balloon at the bursting point is found using the combined gas law, which states that the ratio of the product of pressure and volume to temperature remains constant for a fixed amount of gas. The helium balloon starts at a volume of 5.44 L and a pressure of 2.46 atm, and it will burst when the external pressure reaches 440.0 mmHg. To find the volume of the balloon at this pressure, we need to apply the following transformation using the combined gas law:
(Initial Pressure) x (Initial Volume) = (Final Pressure) x (Final Volume)

2.46 atm x 5.44 L = (440.0 mmHg / 760 mmHg/atm) x (Final Volume)

Final Volume = (2.46 atm x 5.44 L) / (440.0 mmHg / 760 mmHg/atm)

Final Volume = (13.3864 atmĀ·L) / (0.5789 atm) = 23.1081 L

The balloon will burst when it reaches a volume of approximately 23.11 L, which is not an option provided in the question. It appears there may be a mistake in the question or in the options provided. The key point is to understand the application of the combined gas law and pressure conversions to solve the problem.

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